The sine inverse of a purely complex number

Member
To prove that
$$\displaystyle sin^{-1}(ix)=2n\pi\pm i log(\sqrt{1+x^2}+x)$$
I can prove $$\displaystyle sin^{-1}(ix)=2n\pi+ i log(\sqrt{1+x^2}+x)$$

but facing problem to prove
$$\displaystyle sin^{-1}(ix)=2n\pi- i log(\sqrt{1+x^2}+x)$$

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ZaidAlyafey

Well-known member
MHB Math Helper
Re: Complex no

To prove that
$$\displaystyle sin^{-1}(ix)=2n\pi\pm i log(\sqrt{1+x^2}+x)$$
I can prove $$\displaystyle sin^{-1}(ix)=2n\pi+ i log(\sqrt{1+x^2}+x)$$

but facing problem to prove
$$\displaystyle sin^{-1}(ix)=2n\pi- i log(\sqrt{1+x^2}+x)$$

Can you sketch your proof ? what is $n$ ?

mathbalarka

Well-known member
MHB Math Helper
Hint : $e^{ix} = \cos(x) + i\sin(x)$.

ZaidAlyafey said:
what is n?
An arbitrary integer of course.

Member
Re: Complex no

Can you sketch your proof ? what is $n$ ?
n is any integer

Let $$\displaystyle sin^{-1}(ix)=\theta$$

Then $$\displaystyle sin\theta = ix$$

$$\displaystyle cos\theta=\pm\sqrt{1-sin^2\theta}=\pm\sqrt{1+x^2}$$

Now $$\displaystyle e^{i\theta}=cos\theta +i sin\theta=\pm\sqrt{1+x^2}+i i x=-x\pm\sqrt{1+x^2}$$

Therefore $$\displaystyle i\theta=Log (-x+\sqrt{1+x^2})(Only + sign ~~considered)=2n\pi i +log(-x+\sqrt{1+x^2})$$

$$\displaystyle \theta = 2n\pi-i log(-x+\sqrt{1+x^2})$$
$$\displaystyle \theta = 2n\pi+i log(x+\sqrt{1+x^2})$$

mathbalarka

Well-known member
MHB Math Helper
In order to get the other sign, simply note that $\sin(-x) = -\sin(x)$.

In order to get the other sign, simply note that $\sin(-x) = -\sin(x)$.